16
$\begingroup$

I am having trouble with Spivak's proof of the Inverse Function Theorem in his Calculus on Manifolds:

2-11 Theorem (Inverse Function Theorem). Suppose that $f: \mathbb{R}^n\to\mathbb{R}^n$ is continuously differentiable in an open set containing $a$, and det $f'(a)\neq 0$. Then there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f:V\to W$ has a continuous inverse $f^{-1}:W\to V$ which is differentiable and for all $y\in W$ satisfies $$(f^{-1})'(y) = [f'(f^{-1}(y))]^{-1}$$ Proof. Let $\lambda$ be the linear transformation $Df(a)$. Then $\lambda$ is non-singular, since det $f'(a)\neq 0$. Now $D(\lambda^{-1}\circ f)(a) = D(\lambda^{-1})(f(a))\circ Df(a) = \lambda^{-1}\circ Df(a)$ is the identity linear transformation. If the theorem is true for $\lambda^{-1}\circ f$, it is clearly true for $f$...

How is the theorem true for $f$ if it is true for $\lambda^{-1}\circ f$?

$\endgroup$
2
  • 1
    $\begingroup$ I have not read Spivak's book, but I would have guessed $\lambda$ is the constant map $x\rightarrow \lambda x$, which does not influence $f$'s differentiablity since $\lambda f$ is just multiplying $f$ by a constant. $\endgroup$ Sep 1, 2012 at 20:35
  • $\begingroup$ I think the key difficulty around this point isn't that $f$ is differentiable if $\lambda$$f$ is but that the formula for the inverse remains true. $\endgroup$ May 27 at 8:50

3 Answers 3

25
$\begingroup$

$\lambda\colon \mathbb{R}^n\to\mathbb{R}^n$ is a bijection and $\lambda$ and $\lambda^{-1}$ are both continuously differentiable. Note that $\lambda'(z) = \lambda$ for all $z \in \mathbb{R}$.

Let $g = \lambda^{-1}\circ f$. Suppose the theorem is true for $g$. Then there is an open set $V'$ containing $a$ and an open set $W'$ containing $g(a)$ such that $g:V'\to W'$ has a continuous inverse $g^{-1}:W'\to V'$ which is differentiable and for all $y\in W'$ satisfies $$(g^{-1})'(y) = [g'(g^{-1}(y))]^{-1}$$ Then $\lambda(W')$ is open and $f = \lambda\circ g:V'\to \lambda(W')$ has a continuous inverse $g^{-1}\circ \lambda^{-1}:\lambda(W')\to V'$.

By the chain rule, for all $z \in \lambda(W')$, $(f^{-1})'(z) = (g^{-1}\circ\lambda^{-1})'(z) = (g^{-1})'(\lambda^{-1}(z))\circ \lambda^{-1} = [g'(g^{-1}(\lambda^{-1}(z)))]^{-1}\circ \lambda^{-1} = [g'(f^{-1}(z))]^{-1}\circ \lambda^{-1} = [\lambda\circ g'(f^{-1}(z)]^{-1} = [f'(f^{-1}(z))]^{-1}$

$\endgroup$
3
  • $\begingroup$ Why is $\lambda$ a bijection? $\endgroup$ May 27, 2018 at 12:07
  • $\begingroup$ $\lambda$ is non-singular (since $\det f'(a) \not= 0$). Hence it estabilish a one-to-one correspondence between the function domain and image (i.e. it is a bijection) $\endgroup$ Mar 28, 2019 at 7:53
  • $\begingroup$ $Df(a)$ is a linear mapping, but $f'(a)$ is a matrix. So, $\lambda^{'}(z)$ is a matrix but $\lambda$ is a linear mapping. So, $\lambda^{'}(z)\neq\lambda$. $\endgroup$
    – tchappy ha
    Aug 27, 2022 at 20:08
9
$\begingroup$

Suppose the theorem holds for all functions $f$ such that $Df(a) = I$. What Spivak has shown is that for $\lambda = Df(a)$, $D(\lambda^{-1} \circ f(a)) = I$. By the assumption above, the theorem must be true for $\lambda^{-1}\circ f$. That means there exists $g$ which is the continuously differentiable inverse of $\lambda^{-1} \circ f$. This implies that $(g \circ \lambda^{-1}) \circ f = I$. Now $g \circ \lambda^{-1}$ is continuously differentiable, and so the theorem is true for $f$.

$\endgroup$
3
  • $\begingroup$ I dont understand the last statement. How does $g \circ \lambda ^{-1} $being continuously differentiable imply the theorem is true for $f$? $\endgroup$ May 27, 2018 at 12:11
  • $\begingroup$ Notice how he placed the parentheses around $g\circ\lambda^{-1}$. This has no effect on computing the compositions; he simply did it to show you that $g\circ\lambda^{-1}$ is in fact the inverse of $f$ that we are looking for! And in addition, this inverse is continuously differentiable since it is just the map $g$ and we generated $g$ as a result of applying the Inverse function Theorem to $\lambda^{-1} \circ f(a)$, so $g$ certainly satisfies all the properties that we'd want from our inverse. $\endgroup$ Jun 7, 2018 at 0:01
  • $\begingroup$ The inverse is continuous and differentiable and may not be continuously differentiable $\endgroup$
    – chesslad
    Sep 23, 2020 at 19:39
1
$\begingroup$

Let $f: \mathbb{R}^n\to\mathbb{R}^n$ be a continuously differentiable function in an open set containing $a$, and det $f'(a)\neq 0$.
Let $\lambda := Df(a)$.
Then $\lambda$ is a non-singular linear transformation because $\det f'(a) \neq 0$.
Let $g := \lambda^{-1} \circ f$ and $f = (f^1, \cdots, f^n)$ and $g = (g^1, \cdots, g^n)$.
Then we can write $g^i(x)$ as $g^i(x) = \mu_{i 1} f^1(x) + \cdots + \mu_{i n} f^n(x)$ for some real numbers $\mu_{i 1}, \cdots, \mu_{i n}$.
Then each $g^i$ is a continuously differentiable function in an open set containing $a$ because each $f^i$ is a continuously differentiable function in an open set containing $a$.
So, $g:\mathbb{R}^n\to\mathbb{R}^n$ is continuously differentiable in an open set containing $a$.

$D(\lambda^{-1} \circ f)(a) = D\lambda^{-1}(f(a)) \circ Df(a) = \lambda^{-1} \circ Df(a) = I$.
So, $g'(a) = I_n$.
So, $\det g'(a)=1\neq 0$.

Then by the Spivak's proof in the text, there is an open set $V$ containing $a$ and an open set $W$ containing $g(a)$ such that $g:V\to W$ has a continuous inverse $g^{-1} : W \to V$ which is differentiable and for all $y\in W$ satisfies $$(g^{-1})^{'}(y)=[g^{'}(f^{-1}(y))]^{-1}.$$

$f = \lambda \circ g$.

The image of $\lambda : W \to \mathbb{R}^n$ is $\lambda (W)$.
$\lambda : W \to \lambda (W)$ is injective because $\lambda : \mathbb{R}^n \to \mathbb{R}^n$ is non-singular.
So, $\lambda : W \to \lambda (W)$ is bijective.
So $f=\lambda\circ g : V \to \lambda (W)$ is bijective.

$\lambda W$ is an open set. A proof of this fact is the following:
Let $\lambda(w) \in \lambda (W)$.
By the problem 1-10 on p.5, there exists a positive real number $M$ such that $|\lambda^{-1} (h)| \leq M |h|$ for any $h \in \mathbb{R}^n$.
Because $W$ is open, there exists $\epsilon > 0$ such that for any $w' \in \mathbb{R}^n$, if $|w' - w| < \epsilon$, then $w' \in W$.
Let $u'$ be an arbitrary element of $\mathbb{R}^n$ such that $|u' - \lambda w| < \frac{\epsilon}{M}$.
Then, $|\lambda^{-1} (u') - w| = |\lambda^{-1} (u') - \lambda^{-1} (\lambda w)| = |\lambda^{-1} (u' - \lambda w)| \leq M |u' - \lambda w| < M \frac{\epsilon}{M} = \epsilon$.
So, $\lambda^{-1} (u') \in W$.
So, $u' = \lambda(\lambda^{-1}(u')) \in \lambda (W)$.
So, $\lambda (W)$ is open.

Now $f : V \to \lambda (W)$ is a continuously differentiable bijective function from an open set $V$ to an open set $\lambda (W)$.

$f^{-1} = g^{-1} \circ \lambda^{-1}$.
$g^{-1}$ is differentiable on $W$ and $\lambda^{-1}$ is differentiable on $\lambda (W)$.
So, $f^{-1}$ is differentiable on $\lambda (W)$ by Chain Rule.
Let $\lambda^{'}(x)=A$.
Then, \begin{align*} (f^{-1})^{'}(y) &= (g^{-1}\circ\lambda^{-1})^{'}(y) = (g^{-1})^{'}(\lambda^{-1}(y))\cdot(\lambda^{-1})^{'}(y)\\ &=[g^{'}(g^{-1}(\lambda^{-1}(y)))]^{-1}\cdot A^{-1} =[g^{'}(f^{-1}(y))]^{-1}\cdot A^{-1}\\ &=[A\cdot g^{'}(f^{-1}(y))]^{-1} =[(\lambda\circ g)^{'}(f^{-1}(y))]^{-1}\\ &=[f^{'}(f^{-1}(y))]^{-1}. \end{align*}

$\endgroup$
1
  • $\begingroup$ I totally agree that this discussion is needed. I've googled a few of these proofs and there is far too little justification of the assertion that we can assume WLOG that the Jacobian is the identity. $\endgroup$ May 27 at 8:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .